Just in case you don't believe that the line AB above is trisected, I can show that it is.

For this, I'm going to rely on Cartesian coordinates. I've chosen not to use standard Euclidian proofs, simply because my geometry is not good enough to do that.

Take point A as the origin of the Cartesian coordinate system, (0,0)

Assume point B lies at (12,0). Call it (12N,0), if you really want to be pernickety.

ABC and ABD are equilateral triangles, so point C is at (6,6√3), while D is at (6,-6√3).

ABDE is a rhombus with side 12*, which means that G is the mid-point of AD. So if D is at (6,-6√3), then G is at (3,-3√3).

* It is easy to prove this using the lengths of the lines and the angles. We know that AE, AD and AB are the same length (they are all radii) while the angles DAE and DAB are both exactly 60°C. This means the two triangles, ABD and ADE are both equilateral, which makes ABDE a rhombus.

The line CG joins C (6,6√3) and G (3,-3√3). The slope of this line is represented by the difference in y coordinates divided by the difference in x coordinates, or (9√3/3), which is 3√3.

So the equation for line CG is y=3√3x + A. Solving for A using the two given points gives A=-12√3.

Thus, CG is represented by y=3√3x - 12√3.

Solving for y=0, or the intercept point I, gives x=4.

By symmetry, J is also 4 units from point B.

Thus we can see that I and J do indeed trisect the line AB.

Congratulations to Major General Panic and his students.

As to whether that also trisects the angle, Elrac is correct: it doesn't.

While the line is accurately trisected, the angle subtended by the line segment IJ at point C are not the same as the angles subtended by AI or BJ.

MGP's suggested technique is to use the trisected line to create a trisected angle. Use the length AB to subtend the relevant angle, and then use the trisections of the line to divide that angle into three parts

While the three angles might look kind of similar, especially where the angle at C is acute, the limitations of this technique emerge when the angle at C is not only obtuse, but close to 180°.

Say you wanted to trisect the angle of 179°, and used MJP's technique. Each of the three angles should be just under 60°.

However, if the point is close to line AB, in order to subtend an angle of 179°, then the middle portion will still subtend an obtuse angle, while the two outer portions will clearly be much less that 60°.